Just as we can curry functions, so we can "curry" homomorphisms and actions.

Given an action $\rho :G\times X\to X$ of group $G$ on set $X$, we can consider what happens if we fix the first argument to $\rho$. Writing $\rho \left(g\right)$ for the induced map $X\to X$ given by $x\mapsto \rho (g,x)$, we can see that $\rho \left(g\right)$ is a bijection.

Indeed, we claim that $\rho \left(g^{-1}\right)$ is an inverse map to $\rho \left(g\right)$. Consider $\rho \left(g^{-1}\right)\left(\rho \right(g\left)\right(x\left)\right)$. This is precisely $\rho \left(g^{-1}\right)\left(\rho \right(g,x\left)\right)$, which is precisely $\rho (g^{-1},\rho (g,x\left)\right)$. By the definition of an action, this is just $\rho (g^{-1}g,x)=\rho (e,x)=x$, where $e$ is the group's identity.

We omit the proof that $\rho \left(g\right)\left(\rho \right(g^{-1}\left)\right(x\left)\right)=x$, because it is nearly identical.

That is, we have proved that $\rho \left(g\right)$ is in $Sym\left(X\right)$, where $Sym$ is the symmetric group; equivalently, we can view $\rho$ as mapping elements of $G$ into $Sym\left(X\right)$, as well as our original definition of mapping elements of $G\times X$ into $X$.

$\rho$ is a homomorphism in this new sense

It turns out that $\rho :G\to Sym\left(X\right)$ is a homomorphism. It suffices to show that $\rho \left(gh\right)=\rho \left(g\right)\rho \left(h\right)$, where recall that the operation in $Sym\left(X\right)$ is composition of permutations.

But this is true: $\rho \left(gh\right)\left(x\right)=\rho (gh,x)$ by definition of $\rho \left(gh\right)$; that is $\rho (g,\rho (h,x\left)\right)$ because $\rho$ is a group action; that is $\rho \left(g\right)\left(\rho \right(h,x\left)\right)$ by definition of $\rho \left(g\right)$; and that is $\rho \left(g\right)\left(\rho \right(h\left)\right(x\left)\right)$ by definition of $\rho \left(h\right)$ as required.